Problem: Solve the equation. $\dfrac{dy}{dx}=-\dfrac{1}{4}e^x y^{-2}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\pm\sqrt[3]{-\dfrac{3e^x}{4}+C}$ (Choice B) B $y=\sqrt[3]{-\dfrac{3e^x}{4}+C}$ (Choice C) C $y=\pm\sqrt[3]{-\dfrac{3e^x}{4}}+C$ (Choice D) D $y=\sqrt[3]{-\dfrac{3e^x}{4}}+C$
Solution: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=-\dfrac{1}{4}e^x y^{-2} \\\\ \dfrac{dy}{dx}&=-\dfrac{e^x}{4y^2} \\\\ -4y^2\,dy&=e^{x}\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} -4y^2\,dy&=e^{x}\,dx \\\\ \int -4y^2\,dy&=\int e^{x}\,dx \\\\ -\dfrac{4y^3}{3}&=e^x+C_1 \\\\ y^3&=-\dfrac{3e^x}{4}+C \\\\ y&=\sqrt[3]{-\dfrac{3e^x}{4}+C} \end{aligned}$ [Where did we get C?] Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\sqrt[3]{-\dfrac{3e^x}{4}+C}$